How to solve this trigonometric integral $\int \sin^2t\cos^2t\,dt$? $$\int \sin^2t\cos^2t\,dt$$
Since both exponent are pair and $\ge 2$, according to my understanding I should use one of these equality to solve :


*

*$\sin^2t = \dfrac{1-\cos2t}{2}$

*$\cos^2t = \dfrac{1+\cos2t}{2}$

*$\sin t\cos t = \dfrac{\sin2t}{2}$


I've tried replacing with each of these tree, but I am unable to solve.

What is wrong with my understanding ? Maybe I need to integrate by part after (I also tried but was unable to solve again) ? Or maybe the equality I've chosen are wrong ? 
 A: \begin{align}
& \int \frac{\sin^2 2t}{4} dt \\[6pt]
= {} & \int \dfrac{1 - \cos 4t}{8} dt \\[6pt]
= {} & \frac{t}{8} - \frac{\sin 4t}{32} + C
\end{align}
Here $C$ is the constant of indefinite integral.  
A: Hint:
$$\int \sin^2t\cos^2tdt = \int \sin^2t(1-\sin^2t)dt = \int (\sin^2t - \sin^4t)dt$$
Another perspective: 
$$\int \sin^2t\cos^2tdt = \int \frac{\sin^2(2t)}{4}dt =  \int \dfrac{1 - \cos 4t}{8} dt $$
Try either of these approaches; second one is easier to get the answer with.
A: Hint
$$\sin^2(t)\cos^2(t)=\big(\sin(t)\cos(t)\big)^2=\frac 14 \sin^2(2t)=\frac 18\big(1-\cos(4t)\big)$$
I am sure that you can take from here.
A: First notice that:
$$\int \sin^2(t) \cos^2(t) dt = \frac{1}{4} \int \sin^2(2t) dt.$$
Now let $ x = 2t$, so $\frac{dx}{2} = dt$. Now you have:
$$ \frac{1}{8} \int \sin^2(x) dx.$$
The problem is now reduced to calculate the integral of $\sin^2(x)$. Integrating by parts gives:
\begin{align}
\int \sin^2(x) dx &= -\cos(x) \sin(x) + \int \cos^2(x) dx \\
\int \sin^2(x) dx &= -\cos(x) \sin(x) + \int 1 -\sin^2(x) dx \\
\int \sin^2(x) dx &= \frac{1}{2} \left( -\cos(x) \sin(x) + x \right)
\end{align}
Now you only have to substitute the $x$ and multiply the $\frac{1}{8}$.
